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Python: create sublist without copying

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python

I have a question about how to create a sublist (I hope this is the right term to use) from a given list without copying.

It seems that slicing can create sublists, but does it with copying. Here is an example.

In [1]: a = [1,2,3]

In [2]: id(a)
Out[2]: 4354651128

In [3]: b = a[0:2]

In [4]: b
Out[4]: [1, 2]

In [5]: id(b)
Out[5]: 4354621312

In [6]: id(a[0:2])
Out[6]: 4354620880

See here the id of b and a[0:2] are different, although their values are the same. To double check, change the value in a, the value in b does not change.

In [7]: a[1] = 4

In [8]: a
Out[8]: [1, 4, 3]

In [9]: b
Out[9]: [1, 2]

So to get back to my question, how can I create sublists but without copying? I mean, when value of a[1] is set to 4, b will be [1, 4].

I searched around and did not find much help (maybe I am not using the right keywords). Thank you!


Edits:

Thank you all for your comments and answers! Here is what I have learned.

  • There is no built-in way in Python to create a view of a list (or to create a sublist without copying).
  • The easiest way to do this is to use the numpy array.
  • Although numpy array has limitations on data type compared with list, it does serve my purpose (to implement quicksort with no extra memory)

Here is the same process with numpy array.

In [1]: import numpy as np

In [2]: a = np.arange(1,4)

In [3]: a
Out[3]: array([1, 2, 3])

In [4]: b = a[0:2]

In [5]: b
Out[5]: array([1, 2])

In [6]: id(b)
Out[6]: 4361253952

In [7]: id(a[0:2])
Out[7]: 4361254032

In [8]: a[1] = 4

In [9]: a
Out[9]: array([1, 4, 3])

In [10]: b
Out[10]: array([1, 4])
like image 730
Ethan Avatar asked Feb 05 '15 22:02

Ethan


People also ask

Does Python slicing create a copy?

The short answer. Slicing lists does not generate copies of the objects in the list; it just copies the references to them. That is the answer to the question as asked.

Does list slicing create a new list?

List slicing returns a new list from the existing list.


4 Answers

numpy's array objects support this notion of creating interdependent sub-lists, by having slicing return views rather than copies of the data.

Altering the original numpy array will alter the views created from the array, and changes to any of the views will also be reflected in the original array. Especially for large data sets, views are a great way of cutting data in different ways, while saving on memory.

>>> import numpy as np
>>> array1 = np.array([1, 2, 3, 4])
>>> view1 = array1[1:]
>>> view1
array([2, 3, 4])
>>> view1[1] = 5
>>> view1
array([2, 5, 4])
>>> array1
array([1, 2, 5, 4]) # Notice that the change to view1 has been reflected in array1

For further reference, see the numpy documentation on views as well as this SO post.

like image 93
zehnpaard Avatar answered Oct 06 '22 03:10

zehnpaard


There is no way to do this with built in Python data structures. However, I created a class that does what you need. I don't guarantee it to be bug-free, but it should get you started.

from itertools import islice

class SubLister(object):
    def __init__(self, base=[], start=0, end=None):
        self._base = base
        self._start = start
        self._end = end

    def __len__(self):
        if self._end is None:
            return len(self._base) - self._start
        return self._end - self._start

    def __getitem__(self, index):
        self._check_end_range(index)
        return self._base[index + self._start]

    def __setitem__(self, index, value):
        self._check_end_range(index, "list assignment index out of range")
        self._base[index + self._start] = value

    def __delitem__(self, index):
        self._check_end_range(index, "list assignment index out of range")
        del self._base[index + self._start]

    def __iter__(self):
        return islice(self._base, self._start, self._end)

    def __str__(self):
        return str(self._base[self._start:self._end])

    def __repr__(self):
        return repr(self._base[self._start:self._end])

    # ...etc...

    def get_sublist(self, start=0, end=None):
        return SubLister(base=self._base, start=start, end=end)

    def _check_end_range(self, index, msg="list index out of range"):
        if self._end is not None and index >= self._end - self._start:
            raise IndexError(msg)

Example:

>>> from sublister import SubLister
>>> base = SubLister([1, 2, 3, 4, 5])
>>> a = base.get_sublist(0, 2)
>>> b = base.get_sublist(1)

>>> base
[1, 2, 3, 4, 5]
>>> a
[1, 2]
>>> b
[2, 3, 4, 5]
>>> len(base)
5
>>> len(a)
2
>>> len(b)
4

>>> base[1] = 'ref'
>>> base
[1, 'ref', 3, 4, 5]
>>> a
[1, 'ref']
>>> b
['ref', 3, 4, 5]
like image 27
mVChr Avatar answered Oct 06 '22 01:10

mVChr


you can't if you slice a to get b.

All slice operations return a new list containing the requested elements. This means that the following slice returns a new (shallow) copy of the list [1]

[1] https://docs.python.org/2/tutorial/introduction.html

like image 33
Kamyar Ghasemlou Avatar answered Oct 06 '22 01:10

Kamyar Ghasemlou


There is no built-in way to do this. You could create your own list-like class that takes a reference to a list and reimplements all of the list accessor methods to operate on it.

like image 28
tdelaney Avatar answered Oct 06 '22 01:10

tdelaney